Question

giúp mình câu d với ạ

Answer 1

`d)\sqrt{2+\sqrt{2+\sqrt{x}}}=2`       `ĐK: x >= 0`

`=>2+\sqrt{2+\sqrt{x}}=4`

`=>\sqrt{2+\sqrt{x}}=2`

`=>2+\sqrt{x}=4`

`=>\sqrt{x}=2`

`=>x=4` (t/m)

Vậy `x=4`

Answer 2

d) Điều kiện: \(x\ge0\)

\(\sqrt{2+\sqrt{2+\sqrt{x}}}=2\Rightarrow2+\sqrt{2+\sqrt{x}}=4\)

\(\Rightarrow\sqrt{2+\sqrt{x}}=2\Rightarrow2+\sqrt{x}=4\)

\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(TM\right)\)

Answer 3

`a)`

`| x-36|=1.4`

`<=> x-36 = 1.4` hoặc `x-36=-1,4`

`<=> x = 37,4` hoặc `x = 34,6`

Vậy `...`

`b)`

`1/( x+3)=5/4`

`<=> 5/(5(x+3)) = 5/4`

`=> 5( x + 3 ) = 4`

`<=> 5x+15=4`

`<=> 5x=-11`

`<=> x = -11/5`

Vậy `...`

`c)`

`1/(1.3)+1/(3.5)+....+1/(x(x+2)) = 8/17`

`<=>` `2/(1.3)+2/(3.5)+....+2/(x(x+2)) = 16/17`

`<=> 1 - 1/3 + 1/3 - 1/5 +....+1/x - 1/(x+2) = 16/17`

`<=> 1 - 1/(x+2) = 16/17`

`<=> 1/(x+2)=1/17`

`=>x+2=17`

`<=> x=15`

Vậy `....`

`d)`

\(\sqrt{2+\sqrt{2+\sqrt{x}}}=2\)

`<=> 2 +` \(\sqrt{2+\sqrt{x}}=4\)

`<=>` \(\sqrt{2+\sqrt{x}}=2\) 

`<=>` \(2+\sqrt{x}=4\)

`<=>` \(\sqrt{x}=2\)

`<=> x=4`

Vậy `...`